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\(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}=\frac{12}{25}+\frac{15}{25}+\frac{13}{25}=\frac{12+15+13}{25}=\frac{40}{25}=\frac{8}{5}\)
\(1,\dfrac{12}{25}+\dfrac{3}{5}+\dfrac{13}{25}=\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{3}{5}=1+\dfrac{3}{5}=\dfrac{8}{5}\\ 2,\dfrac{3}{2}+\dfrac{2}{3}+\dfrac{4}{3}=\dfrac{3}{2}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{3}{2}+2=\dfrac{7}{2}\\ 3,\dfrac{3}{2}+\dfrac{7}{5}+\dfrac{3}{4}=\left(\dfrac{3}{2}+\dfrac{3}{4}\right)+\dfrac{7}{5}=\left(\dfrac{6}{4}+\dfrac{3}{4}\right)+\dfrac{7}{5}=\dfrac{9}{4}+\dfrac{7}{5}=\dfrac{45}{20}+\dfrac{28}{20}=\dfrac{73}{20}\)
12/35 + 3/5 + 13/25
= ( 12/25 + 13/25 ) + 3/5
= 1 + 3/5
= 8/5
\(a.\dfrac{12+x}{42}=\dfrac{35}{42}\Leftrightarrow12+x=35\Leftrightarrow x=23\)
\(b.\dfrac{25-x}{40}=\dfrac{3}{8}\Leftrightarrow\dfrac{25-x}{40}=\dfrac{15}{40}\Leftrightarrow25-x=15\Leftrightarrow x=10\)
\(c.\dfrac{13+x}{20}=\dfrac{3}{4}\Leftrightarrow\dfrac{13+x}{20}=\dfrac{15}{20}\Leftrightarrow13+x=15\Leftrightarrow x=2\)
\(d.\dfrac{23-x}{25}=\dfrac{20}{25}\Leftrightarrow23-x=20\Leftrightarrow x=3\)
Tính thuận tiện :
\(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
= \(\frac{12}{25}+\frac{13}{25}+\frac{3}{5}\)
= \(\frac{25}{25}+\frac{3}{5}\)
= \(1+\frac{3}{5}\)
= \(\frac{5}{5}+\frac{3}{5}\)
= \(\frac{8}{5}\)
\(\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
\(=\left(\frac{12}{25}+\frac{13}{25}\right)+\frac{3}{5}\)
\(=\frac{25}{25}+\frac{3}{5}\)
\(=1+\frac{3}{5}=\frac{8}{5}\)
\(a,\frac{12}{25}+\frac{3}{5}+\frac{13}{25}\)
\(=\left(\frac{12}{25}+\frac{13}{25}\right)+\frac{3}{5}\)
\(=\frac{25}{25}+\frac{3}{5}\)
\(=1+\frac{3}{5}\)
\(=\frac{8}{5}\)
\(b,\frac{3}{2}+\frac{2}{3}+\frac{4}{3}\)
\(=\frac{3}{2}+\left(\frac{2}{3}+\frac{4}{3}\right)\)
\(=\frac{3}{2}+2\)
\(=\frac{7}{2}\)
\(c,\frac{3}{5}+\frac{7}{5}+\frac{3}{4}\)
\(=\left(\frac{3}{5}+\frac{7}{5}\right)+\frac{3}{4}\)
\(=2+\frac{3}{4}\)
\(=\frac{11}{4}\)
12/25 + 3/5 + 13/25
= 12/25 + 13/25 + 3/5
= 1 + 3/5
= 8/5